Quantitative Aptitude: Quadratic Equations Set 24

Quadratic Equations Practice Sets for IBPS PO, NICL, NIACL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive exams.

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

  1. I. 5x2  – 18x + 9 =0
    II. 3y2 + 5y – 2 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
     Option A
    Solution:

    5x2   – 18x + 9  = 0
    => 5x2  – 15x – 3x  + 9 =0
    => (5x – 3 )(x-3 )= 0
    => x=  3/ 5 or  x= 3
    3y2 + 5y – 2 = 0
    => 3y2  + 6y – y -2 = 0
    => (3y-1)(y + 2) = 0
    => y = 1/3  or  -2
  2. I. √x  – √6 / √x = 0
    II. y3 – 63/2  = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
     Option E
    Solution: 

    √x  –   √6 / √x = 0
    x – √6 = 0
    x = √6
    y3   – 6 (3/2 )  = 0
    =>y3  = (√6)3
    => y = √6
  3. I. (625)1/4x + √1225  = 155
    II. √196y + 13  = 279
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
     Option A
    Solution: 

    5x + 35 = 155
    => 5x = 155 – 35
    => x = 120/ 5 = 24√196 y  + 13 = 279
    => 14y = 279 -13
    => y = 266/14 =19
  4. I. 3x2 – 17x + 24 =0
    II. 4y2  – 15y + 14 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
     Option A
    Solution: 

    3x2  -17x  + 24 =0
    => 3x2 – 9x  – 8x +24 = 0
    => (3x- 8)(x-3) = 0
    =>  x=  8/3 or 3
    4y2  – 15y + 14 = 0
    => 4y2  – 8y -7y +14 = 0
    => (4y -7 )(y-2)=0
    => y = 7 /4 or  2
  5. I. x2 – 2x – √5 x + 2√5 = 0
    II. y2 – √3 y – √2 y + √6 =0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
     Option A
    Solution: 

    x2  – 2x- √5 x + 2√5 = 0
    => x(x-2 ) – √5 (x-2 )= 0
    => (x-2)(x-√5)=0
    => x= 2 0r √5
    y2  – √3 y  – √2 y + √6 =0
    => y(y-√3) – √2(y – √3) = 0
    => (y – √2)(y- √3) =0
    => y = √2 or √3
  6. I. 5x2 – 23x + 12 = 0,
    II. 5y2 – 28y + 15 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
     Option E
    Solution: 

    x = 3/5, 4
    y = 3/5, 5
  7. I. 6x2 + 5x – 6 = 0,
    II. 3y2 – 11y + 6 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option D
    Solution:

    x = 2/3, 3
    y = -3/2, 2/3
  8. I. 3x2 – 5x – 12 = 0,
    II. 2y2 – 17y + 36 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option B
    Solution:

    x = -4/3, 3
    y = 4, 9/2

  9. I. 8x2 – (4 + 4√3)x + 2√3 = 0,
    II. 3y2 – (4 + 3√3)y + 4√3 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    Option B
    Solution:

    8x2 – (4 + 4√3)x + 2√3 = 0
    (8x2 – 4x) – (4√3x – 2√3) = 0
    4x (2x- 1) – 2√3 (2x – 1) = 0,
    So x = 1/2 (0.5), 2√3/4 (0.87)
    3y2 – (4 + 3√3)y + 4√3 = 0
    (3y2 – 4y) – (3√3y – 4√3) = 0
    y (3y – 4) – √3 (3y – 4) = 0
    So, y = √3 (1.732), 4/3
    put on number line
    0.5………..0.87……..4/3………1.732
  10. I. x2 + (4 + 2√2)x + 8√2 = 0
    II. 3y2 – (3 + √3)y + √3 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option B
    Solution:

    x2 + (4 + 2√2)x + 8√2 = 0
    (x2 + 4x) + (2√2x + 8√2) = 0
    x (x + 4) + 2√2 (x + 4) = 0
    So x = -4, -2√2 (-2.8) 
    3y2 – (3 + √3)y + √3 = 0
    (3y2 – 3y) – (√3y – √3) = 0
    3y (y – 1) – √3 (y – 1) = 0
    So y = 1, √3/3 (0.57)

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25 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 24”

  1. Pawan

    please check ques 1 its coming option E

      1. thinker

        Mam question 6me ,x=3/5,4 and y=3/5,5 to eska answer no relation kese Hua .Y bda ya equal hona chahiye.ye question mujhse hmesha wrong ho hate h.why no relation?

        1. ครђ

          Value of X= 3/5 , 4
          Value of Y= 3/5 , 5
          Compare 4 & 3/5 we get x>y …4 and 5 x<y

      2. Pawan

        value of x is same but value of y coming 2/3,-1 please check my eq can i make it as 3y^2+3y+2y-2=0

        1. ครђ

          Value of Y = -6/3 & 1/3
          = -2 ,0.3
          Value of X = 3, 0.6
          So X>Y option A 🙂

  2. Pawan

    also check ques 5
    ans should b option (a)

  3. Barish!!

    thanku mam /

  4. Jellyfish

    8/10 Thx az
    Queno 5 option should be A
    Que 6 2nd statement 5y^2-28y+15=0 he chahiye

  5. Sachin Shukla

    10/10 ty:)))

  6. Sindabad

    @s@Shubhra_AspirantsZone:disqus sir, are u same subhra sir from Affairscloud??

    1. Yes the same

      And i am female 🙂

      1. Sindabad

        Sorry, good morning ma’am

  7. swati

    thank you team 10/10 easy….provide more…….sets like this……

  8. anurag agarwal

    Unable to solve question 5.9.10
    How to solve it

    1. What doubt ?
      See 9th solution.
      U have to take without root values together
      And with root s values together
      And then take common like in 2nd step in 9th.

      See again. Ask if doubt

  9. Ayushi

    thankuu mam 🙂

  10. Chotu D(mind) !! {NONSENSE}

    done

  11. वैम्पायर हंटर

    6th ka sign chk kro

      1. वैम्पायर हंटर

        Wo 6th ke 2nd equtn me sign minus minus hh
        To fr value dono plus me likh rkhe ho ye bol rha hun…

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